Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.1.44c

Chapter 7, Problem 7.1.44c

In Exercises 41–44:

c. Evaluate df/dx at x = a and df⁻¹/dx at x = f(a) to show that
(df⁻¹/dx)|ₓ₌f(a) = 1 / (df/dx)|ₓ₌a

44. f(x) = 2x², x ≥ 0, a = 5

Verified step by step guidance

First, identify the function and the point at which you need to evaluate the derivatives. Here, the function is \(f(x) = 2x^{2}\) with \(x \geq 0\), and the point is \(a = 5\).

Calculate the derivative of \(f(x)\) with respect to \(x\). Use the power rule: \(\frac{d}{dx} (2x^{2}) = 2 \cdot 2x^{2-1} = 4x\). So, \(\frac{df}{dx} = 4x\).

Evaluate \(\frac{df}{dx}\) at \(x = a = 5\). This gives \(\left. \frac{df}{dx} \right|_{x=5} = 4 \times 5\).

Next, find the inverse function \(f^{-1}(x)\). Since \(f(x) = 2x^{2}\) and \(x \geq 0\), solve for \(x\) in terms of \(y\): \(y = 2x^{2} \Rightarrow x = \sqrt{\frac{y}{2}}\). Thus, \(f^{-1}(x) = \sqrt{\frac{x}{2}}\).

Calculate the derivative of the inverse function \(f^{-1}(x)\) with respect to \(x\). Using the chain rule, \(\frac{d}{dx} \left( \sqrt{\frac{x}{2}} \right) = \frac{1}{2} \left( \frac{x}{2} \right)^{-\frac{1}{2}} \cdot \frac{1}{2} = \frac{1}{4} \left( \frac{x}{2} \right)^{-\frac{1}{2}}\). Then, evaluate this derivative at \(x = f(a) = f(5) = 2 \times 5^{2}\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative of a Function

The derivative of a function f(x) at a point x = a represents the instantaneous rate of change or slope of the function at that point. It is found by taking the limit of the difference quotient as the interval approaches zero. For example, if f(x) = 2x², then df/dx = 4x, and at x = 5, df/dx = 20.

Inverse Function and Its Derivative

An inverse function f⁻¹(x) reverses the effect of f(x), so that f(f⁻¹(x)) = x. The derivative of the inverse function at a point x = f(a) relates to the derivative of the original function at x = a. Specifically, (df⁻¹/dx) at x = f(a) equals the reciprocal of (df/dx) at x = a, provided the derivative is nonzero.

Chain Rule and Its Application to Inverse Functions

The chain rule states that the derivative of a composite function is the product of the derivatives of the inner and outer functions. Applying this to f(f⁻¹(x)) = x, differentiating both sides yields (df/dx)|ₓ₌f⁻¹(x) * (df⁻¹/dx) = 1. This relationship is key to proving that the derivative of the inverse is the reciprocal of the original function's derivative.

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Related Practice

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In Exercises 41–44:c. Evaluate df/dx at x = a and df⁻¹/dx at x = f(a) to show that(df⁻¹/dx)|ₓ₌f(a) = 1 / (df/dx)|ₓ₌a44. f(x) = 2x², x ≥ 0, a = 5

Verified video answer for a similar problem:

Key Concepts

Derivative of a Function

Inverse Function and Its Derivative

Chain Rule and Its Application to Inverse Functions

In Exercises 41–44:

c. Evaluate df/dx at x = a and df⁻¹/dx at x = f(a) to show that
(df⁻¹/dx)|ₓ₌f(a) = 1 / (df/dx)|ₓ₌a

44. f(x) = 2x², x ≥ 0, a = 5